Menu item that sets properties on menu open

Question

I currently trying to create a dynamically created context menu. I'm currently binding a ObservableCollection<MenuItem> to the ItemsSource property on the context menu. I now want to set the visibility of the items in the list when the menu opens depending on what I have selected.
I tried Inheriting from MenuItem like this

public class CtContextMenuItem : MenuItem { public delegate Visibility VisibilityDelegate(); public VisibilityDelegate IsVisibleDelegate = null; }

And I want to set Visibility to the result of the VisibilityDelegate when the context menu is opened but I can't find any event or method that is called on the MenuItem when the context menu is opened
Is there a way to do it or is it do I have to just create all the items of the menu inside a function listening to ContextMenuOpening?

"listening to ContextMenuOpening" - yes. Enumerate menu items and call that delegate either directly or via property (if data template binding is used). Consider to add a little of mvvm instead of using custom controls. — Sinatr
– Sinatr, Commented Oct 10, 2017 at 9:17

mm8 · Accepted Answer · 2017-10-10 10:24:43Z

Bind the ItemsSource to an ObservableCollection<CtContextMenuItem> where the CtContextMenuItem type has a Visibility or bool property that you can bind to in your XAML. Something like this:

public class CtContextMenuItem { public Visibility IsVisible { get; set; } }

<ContextMenu ItemsSource="{Binding TheSourceCollection}"> <ContextMenu.ItemContainerStyle> <Style TargetType="MenuItem"> <Setter Property="Visibility" Value="{Binding IsVisible}" /> </Style> </ContextMenu.ItemContainerStyle> </ContextMenu>

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